L1. L2 =>  regular . CFL  == CFL. CFL  (as every regular is CFL so we can assume regular as CFL)

Since CFL is closed under concatenation so 

CFL. CFL= CFL  

Hence

Regular . CFL = CFL is true

 

L1 ∪ L2 => Regular ∪ CFL = CFL 

Regular ∪ CFL = CFL ∪ CFL  (as every regular is CFL so we can assume regular as CFL)

 

Since CFL is closed under union 

Hence Regular ∪ CFL = CFL   is true

 

L1 ∩ L2 => Regular ∩ CFL = CFL 

Regular languages are closed under intersection with any language

I.e,

Regular ∩ L = L  (where L is any language such as CFL, CSL etc)

 

Hence Regular ∩ CFL = CFL  is true

 

Please note this is a special property of regular languages so we will not upgrade regular into CFL (as we did in S1 and S2). We can directly use these closure properties.

 

L1 − L2 => Regular − CFL = CFL 

=> Regular − CFL = regular ∩ CFL (complement)

Since CFL is not closed under complement so complement of CFL may or may not be CFL

 

Hence Regular − CFL need not be CFL

 

For ex:

R= (a+b+c)*  and L= {am bn ck | m ≠ n or n ≠ k} which is CFL.

 

The complement of L = {an bn cn | n>0} which is CSL but not CFL.

 

So 

R − L = (a+b+c)* − {am bn ck | m ≠ n or n ≠ k}

 

=> (a+b+c)* ∩  L (complement) 

=> (a+b+c)* ∩  {an bn cn | n>0}

=> {an bn cn | n>0}

 

Which is CSL. Hence Regular − CFL need not be CFL.

1 ⇒ is recursive,
This one is true, because L₁ is context free which is nothing but recursive, recursive language is closed under complement hence true.
2 ⇒ (complement of L₂) is recursive
If L₂ and both are recursive enumerable then is recursive.
Hence option 2 is false
3 ⇒ is context free
Which is false because context free language does not closed under complement
4 ⇒ ∪L₂ is recursive enumerable
⇒ recursive
Every recursive language is also recursive enumerable
L₂ ⇒ recursive enumerable
∪ L₂ ⇒ recursive enumerable
Because recursive enumerable language closed under union.